There are two electrons inside an infinite potential cube well (essentially $V=\infty$ outside a cube and $0$ inside it) and I need to find the degeneracy of the first excited state. I'm thinking it would be 9 (3 quantum numbers in $x, y, z$ and 3 ways to configure spin to get antisymmetric wavefunction?) but how do I actually get to the right answer?
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1$\begingroup$ This looks like something that belongs in the homework section. For an introductory text on quantum mechanics and multiparticle wavefunctions, I can refer you to D.J. Griffiths' Introduction to Quantum Mechanics, section 5.1. $\endgroup$– JeroenOct 13, 2016 at 17:22
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$\begingroup$ This is exactly the textbook I'm using, and trust me I've read that part multiple times but it isnt making me any smarter or I wouldn't be posting here in the first place :( $\endgroup$– user3411693Oct 13, 2016 at 17:25
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In the first excited state, one of the electrons is excited while the other one is in the ground state. Since the overall wavefunction has to be anti-symmetric it has to be anti-symmetric in the spatial part (since one of the electrons is in the ground state while the other is excited). It furthermore has to be symmetric in the spin part (if it would be anti-symmetric in both the wavefunction and the spin, the total wavefunction would again be symmetric and it would equal zero).
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$\begingroup$ Thank you! So the spin must be symmetric, got it - doesn't that mean that there are 3 different spin states to choose from though, the triplet states? Maybe that doesn't matter? $\endgroup$ Oct 13, 2016 at 17:40
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$\begingroup$ Does the amount of spin states to choose from not affect the degeneracy or am I wrong in assuming there are 3 states to choose from? $\endgroup$ Oct 13, 2016 at 17:56
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$\begingroup$ As you said, there are 3 different antisymmetric wavefunctions and 3 different symmetric spin states. The total number of possible states is thus $3 \times 3 = 9$. $\endgroup$– JeroenOct 13, 2016 at 17:58
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$\begingroup$ So the degeneracy is indeed 9? Thank you for your time $\endgroup$ Oct 13, 2016 at 17:59